Ncert Solutions Chemistry Class 11th

In electro-refining, impure metal (blister copper) is used as an anode while precious metal like Au, Pt gets deposited as anode mud.

$\Delta H=3{\left(\Delta H_{comb}\right)}_{C_2{H}_2\left(g\right)}-{\left(\Delta H_{comb}\right)}_{C_6{H}_6\left(l\right)}$

$=-1300*3-\left(-3268\right)kJ/mole$

$=\left(-3900+3268\right)kJ/mole$

$=-632kJ/mole$

In photo-electric effect;

  $\frac{hc}{\lambda }=h{v}_0+\frac{1}{2}m{v}^2$             

  $\frac{6.63*10^{-34}*3*10^8}{500*10^{-9}}=\left(6.63*10^{-34}*4.3*10^{14}\right)+\frac{1}{2}*9.0*10^{-31}*v^2$            

  $v=\sqrt[]{\frac{\left(3.978*10^{-19}-2.85*10^{-19}\right)}{4.5*10^{-31}}}=\sqrt[]{25*10^{10}}=5*10^{5}m/s$             

Ans. = 5

Hydroxyapatite is the calcium minerals whose molecular formula is $3C{a}_3{\left(P{O}_4\right)}_2.Ca{\left(OH\right)}_2$ ${}_{}{{}_{}}_{}{}_{}$ or $C{a}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2.$ ${}_{}{{}_{}}_{}{}_{}$ During contact with water dissolved F ion, hydroxyapatite converted into ${}_{}{{}_{}}_{}{}_{}$ $3C{a}_3{\left(P{O}_4\right)}_2.Ca{F}_2$ .

$\Delta H_{vap}=41KJmo{l}^{-1}$

$\Delta H=\Delta E+\Delta nRT$      

$H_2{O}_{\left(\mathcal{l}\right)}\to H_2{O}_{vap}$        

$\Delta n=1$             

R = 8.3 JK^-1 mol^-1

T = 373 K

$\Delta$ E (Change of internal energy) = $\Delta H-\Delta nRT$  

$?\Delta E=41-\frac{1*8.3*373}{1000}kJmo{l}^{-1}=41-3.09kJmo{l}^{-1}=37.91kJmo{l}^{-1}$  

Nearest integer = 38.

Statement I is true because methyl orange indicator has pH range (3.2 to 4.2) which is suitable for buffer produced by strong acid and weak base.

Statement II is false because phenolphthalein has pH range (8.4 to 10.2) and only suitable for equivalence point above than 7, In this case pH increased by the addition of NaOH on acetic acid.

De-ionised water means dissolved minerals free water which is prepared by synthetic resin method. Where cation exchanged by H⁺ and anion exchanged by OH^- of resin.

$Ca{\left(OH\right)}_2+C{O}_2\to CaC{O}_3+H_{2}O$ (1st step of reaction)

$CaC{O}_3+H_{2}O+C{O}_2\to Ca{\left(HC{O}_3\right)}_2$ (2nd step of reaction)

In alkaline earth metals cationic size increases down the groups, hence thermal stability of carbonates increases down the group. Therefore both assertion and reason are correct and reason is correct explanation of assertion.

$P=\frac{nRT}{V}=\frac{\left(1+2\right)*8.3*400}{4.0*10^{?3}}=24.9*10^{5}Pa$